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Let $S$ be the set of all $\mathbb Q$-Cauchy sequences. For $(a_n)\in S$ let $[a_n]=\{(b_n)\in S:(a_n)\sim(b_n)\}$. Cauchy sequences are equivalent if they have the same limit.

Suppose $(a_n)\sim (b_n)$ and $(c_n)\sim(d_n)$ then 1. $(a_n)+(c_n)\sim (b_n)+(d_n)$ and 2. $(a_n)(c_n)\sim(b_n)(d_n)$

Reflexive: Suppose $(a_n)\sim (b_n)$ then $a_nRb_n$ so $a_n+b_n\sim a_n+b_n$

Symmetric: Suppose $(a_n)\sim (b_n)$ then $a_nRb_n$ so $a_n+b_n\sim b_n+a_n$

Transitive Suppose $(a_n)\sim (b_n)$ and $(c_n)\sim (d_n)$ then $a_nRb_n$ and $c_nRd_n$, I am not sure what to do with this one?

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    $\begingroup$ what are you trying to do ? proving that ~ is effectively an equivalence relation or trying to prove that $+,\times$ are well defined operations in $\mathbb R$ identified as $S/\sim$ ? It seems there is a bit of a mix right now... $\endgroup$ – zwim Dec 4 '17 at 16:45
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A better definition of Cauchy-sequence equivalence would be

Two Cauchy sequences are equivalent iff their difference is a null Cauchy sequence.

A $\mathbb Q$-Cauchy sequence $\left(a_n\right)$ is said to be a null sequence iff for all $\varepsilon\in\mathbb Q^+$, there exists $N\in\mathbb N$ such that for all $n\ge N$, $|a_n|<\varepsilon$.

The idea of equivalent Cauchy sequences is to use them to construct the real numbers. It will then be seen that the limits of some Cauchy sequences are real. Before we have the idea of a real number, however, it is better not to jump the gun and talk about limits – we can define limits afterwards.

Then for (1): Let $\varepsilon\in\mathbb Q^+$.

$(a_n)\sim(b_n)\ \implies\ \exists N_1:\forall n\ge N_1,\ |a_n-b_n|<\frac\varepsilon2$ $(c_n)\sim(d_n)\ \implies\ \exists N_2:\forall n\ge N_2,\ |c_n-d_n|<\frac\varepsilon2$

Now let $N=\max\{N_1,N_2\}$ and use the fact that $|(a_n+c_n)-(b_n+d_n)|\le|a_n-b_n|+|c_n-d_n|$ to complete the proof.

For (2), you need to use the fact that Cauchy sequences are bounded. In particular $(a_n),(d_n)$ are bounded, i.e.

$|a_n|\le K_1,|d_n|\le K_2$ for some $K_1,K_2\in\mathbb Q^+$.

Now given $\varepsilon\in\mathbb Q^+$, there exist $N_1,N_2$ such that $|a_n-b_n|<\frac\varepsilon{2K_2}$ for all $n\ge N_1$ and $|c_n-d_n|<\frac\varepsilon{2K_1}$ for all $n\ge N_2$. Then for all $n\ge N=\max\{N_1,N_2\}$.

$|a_nc_n-b_nd_n|=|a_n(c_n-d_n)+d_n(a_n-b_n)|\le K_1\cdot\frac\varepsilon{2K_1}+K_2\cdot\frac\varepsilon{2K_2}=\varepsilon$.

Proof that Cauchy sequences are bounded Let $(a_n)$ be a Cauchy sequence. Taking $\varepsilon=1$, $\exists N$ such that $n\ge N$ $\implies$ $|a_n-a_{N+1}|<1$.

So $\forall n\ge N$, $|a_n|=|(a_n-a_{N+1})+a_{N+1}|\le1+|a_{N+1}|$.

$\therefore$ $|a_n|\le K$ for all $n\in\mathbb N$, where $K=\max\{|a_1|,\ldots,|a_N|,1+|a_{N+1}|\}$.

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    $\begingroup$ Perfect and very well explained. +1 $\endgroup$ – Paramanand Singh Dec 4 '17 at 18:36

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